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October 17, 2003

Peas, Carrots, Beans

My daughter brought home the following problem from her 2nd grade math class:

You have 10 pots in which to plant vegetables. You can plant either peas, carrots or beans in each pot. You could plant 7 peas, 2 carrots and 1 bean plant, or 3 peas, 5 carrots and 2 beans, or … as long as you have at least 1 plant of each type and a total of 10 plants. How many different combinations are there?

This is a wonderful problem … for a slightly older child. My daughter’s answer, “more than 10,” was arrived at by a really noble attempt at brute-force enumeration. For 2nd graders, who have not yet mastered division, the actual solution is tantalizingly beyond their grasp.

Let’s solve a simpler problem first. Say we just want to plant two vegetables: peas and carrots. Line the 10 pots up in a row. We’ll plant all the peas on the left and the carrots on the right, and we’ll put a cardboard divider between them. There are 9 places to put the divider, so there are 9 different combinations of 2 vegetables that can be planted in the 10 pots. For 3 vegetables, we need to put in 2 dividers (peas on the left, carrots in the middle and beans on the right). There are 9 places to put the first divider and 8 places to put the second divider (since we’ve already taken one of the slots it could go in). That sounds like 9×8=729\times 8= 72 possibilities. But we’ve overcounted, since swapping the locations of the two dividers gives us the same configuration of plants. So the correct answer is 9×8/2=369\times 8/2=36 combinations. [More generally, with NN pots and kk types of vegetables, there are (N−1k−1)\scriptsize{\left(\array{N-1\\ k-1}\right)} combinations.]

Maybe next year, I could actually explain this solution to her. Maybe, when she’s older still, she could come up with this solution on her own. But not this year …

So what does the teacher have in mind with this exercise?

Posted by distler at October 17, 2003 11:59 PM

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Re: Peas, Carrots, Beans

Re: Peas, Carrots, Beans

I think there is some ambiguity in the way the teacher formulated the problem. Your solution is using the commonsence assumption that the pots are indistinguishable. This is fine if the “pot is quantum” (sorry for the clumsy pun). If the pots are in fact distinct then as far as i can see the solution is more simple: it is simply 3^10.

The explanation as to why the teacher chose this problem will be clear when your daughter gets back the corrected homework: my guess is that the teacher is incompetent and has in its mind a wrong solution.